Pricing for Kiwi Bubble

Abstract

This project aims to illustrate how businesses can utilize choice data and demographic information to understand customer preferences and maximize its profit by studying segmentation and product substitution that is relevant to competitive pricing policies or product line pricing. This project includes 3 parts: 1) logit model without segmentation, 2) logit model with segmentation, 3) understanding strategic responses when rival lowers its price

Environment

In this project, we assume there are only 2 beverage companies (Kiwi and Mango) in the market. Each one of them sells a single product: Kiwi Regular(KR) and Mango Bubble(MB). Now Kiwi is thinking to launch a new product: Kiwi Bubble(KB). Our goal is to make recommendations on whether Kiwi should launch Kiwi Bubble, and if launch, what prices should the firm set for its 2 products in order to maximize its profit. Given that the unit cost of all three products is $0.5, the market size is 1000 customers.

Part 1 - Multinomial logit model without segmentation

Part 1.1 - Estimate a demand model

We first estimate a demand model to predict individual purchase decisions, utilizing multinomial logit model.

#Estimate multinomial logit model
#Convert data using 'mlogit.data'
mlogitdata = mlogit.data(data, id = 'id', varying = 4:7,choice = 'choice', shape = 'wide')
#Run MLE
mle = gmnl(choice ~ price, data = mlogitdata)
summary(mle) 

## 
## Model estimated on: Tue Feb 18 20:57:04 2020 
## 
## Call:
## gmnl(formula = choice ~ price, data = mlogitdata, method = "nr")
## 
## Frequencies of categories:
## 
##       0      KB      KR      MB 
## 0.41564 0.18035 0.20039 0.20362 
## 
## The estimation took: 0h:0m:0s 
## 
## Coefficients:
##                Estimate Std. Error z-value  Pr(>|z|)    
## KB:(intercept)  4.25316    0.32821  12.959 < 2.2e-16 ***
## KR:(intercept)  4.36240    0.32945  13.241 < 2.2e-16 ***
## MB:(intercept)  4.20440    0.31331  13.419 < 2.2e-16 ***
## price          -3.73793    0.23671 -15.791 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Optimization of log-likelihood by Newton-Raphson maximisation
## Log Likelihood: -1909
## Number of observations: 1547
## Number of iterations: 4
## Exit of MLE: gradient close to zero

Preview of the functions that will be used in later part.

#beta0KB = 4.25316, beta0KR = 4.36240, beta0MB = 4.20440, beta1 = -3.73793 
#beta0 of three products are similar, meaning they are selected equally frequently
beta1 = -3.73793
beta0KB = 4.25316
beta0KR = 4.36240
beta0MB = 4.20440


#Functions:
#Demand:
#Define a demand fuction for when there are only 2 products (Kiwi Regular and Mango Bubble) in the market
para=c(beta0KB,beta0KR,beta0MB,beta1)

demand2=function(price1,price2,para){  #2 products, para contain all 4 numbers
  prob=exp(para[1]+para[4]*price1)/(1+exp(para[1]+para[4]*price1)+exp(para[2]+para[4]*price2))
  return(prob)
}

#Define a demand fuction for when there are 3 products (Kiwi Regular, Kiwi Bubble and Mango Bubble) in the market
demand1=function(priceKB,priceKR,priceMB,para){
  prob=exp(para[1]+para[4]*priceKB)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  return(prob)
}

#Define an aggregate choice fuction for 2 products
agg_choice2=function(priceKB,priceKR, own, rival) {  #2 products
  m=c(own+1,rival+1,5)
  agg_choice=seg.share[1]*demand2(priceKB,priceKR,as.numeric(coef.est[1,m]))+
    seg.share[2]*demand2(priceKB,priceKR,as.numeric(coef.est[2,m]))+
    seg.share[3]*demand2(priceKB,priceKR,as.numeric(coef.est[3,m]))+
    seg.share[4]*demand2(priceKB,priceKR,as.numeric(coef.est[4,m]))+
    seg.share[5]*demand2(priceKB,priceKR,as.numeric(coef.est[5,m]))+ 
    seg.share[6]*demand2(priceKB,priceKR,as.numeric(coef.est[6,m]))+
    seg.share[7]*demand2(priceKB,priceKR,as.numeric(coef.est[7,m]))+
    seg.share[8]*demand2(priceKB,priceKR,as.numeric(coef.est[8,m]))+
    seg.share[9]*demand2(priceKB,priceKR,as.numeric(coef.est[9,m]))
  return(agg_choice)
}

#Define an aggregate choice function for 3 products
agg_choice=function(priceKB,priceKR,priceMB) {
  
  agg_choice=seg.share[1]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[1,2:5]))+
    seg.share[2]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[2,2:5]))+
    seg.share[3]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[3,2:5]))+
    seg.share[4]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[4,2:5]))+
    seg.share[5]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[5,2:5]))+ 
    seg.share[6]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[6,2:5]))+
    seg.share[7]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[7,2:5]))+
    seg.share[8]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[8,2:5]))+
    seg.share[9]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[9,2:5]))
  
  return(agg_choice)
}

#sum demand of KB+KR (based on 3 products environment)
demand_KB_KR=function(priceKB,priceKR,priceMB,para){
  probKB=exp(para[1]+para[4]*priceKB)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  probKR=exp(para[2]+para[4]*priceKR)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  return(cbind(probKB,probKR))
}
demand_KB_KR_MB=function(priceKB,priceKR,priceMB,para){
  probKB=exp(para[1]+para[4]*priceKB)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  probKR=exp(para[2]+para[4]*priceKR)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  probMB=exp(para[3]+para[4]*priceKR)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  return(cbind(probKB,probKR,probMB))
}

#sum profit of KB+KR (based on 3 products environment)
profit_KB_KR=function(priceKB,priceKR, priceMB,para){
  profitKB=1000*(demand_KB_KR(priceKB,priceKR, priceMB,para)[,1]*(priceKB-0.5))
  profitKR=1000*(demand_KB_KR(priceKB,priceKR, priceMB,para)[,2]*(priceKR-0.5))
  return(cbind(profitKB,profitKR))
}

Part 1.2 - Calculate elasticities

Now we calculate the own- and cross-elasticities for all combinations of products with average prices observed.

Insights: We can see from the data frame below that all the 3 products KB, KR and MB are price elastic as their own elasticities > 1.

If we solely consider the cross elasticities when making business decisions, launching KB might not be a smart action since KR and MB are close substitutes of each other, and KR is a close substitute of KB which implies a risk of cannibalization.

#Elasticity

#beta1=-3.73793
#beta0KB = 4.25316
#beta0KR = 4.36240
#beta0MB = 4.20440
avgPriceKB = mean(data$price.KB) #1.381332
avgPriceKR = mean(data$price.KR) #1.378087
avgPriceMB = mean(data$price.MB) #1.345585
avgPriceAll=c(avgPriceKB,avgPriceKR,avgPriceMB)

#KB+KR+MB
demandKB = demand1(avgPriceKB,avgPriceKR,avgPriceMB,para) 
demandKR = demand1(avgPriceKR,avgPriceKB,avgPriceMB,para[c(2,1,3,4)])  
demandMB = demand1(avgPriceMB,avgPriceKB,avgPriceKR,para[c(3,1,2,4)]) 
#Own-price elasticity of KB
ownElasKB = -(beta1)*avgPriceKB*(1-demandKB) #4.222673
#Own-price elasticity of KR
ownElasKR = -(beta1)*avgPriceKR*(1-demandKR) #4.143605
#Own-price elasticity of MB 
ownElasMB = -(beta1)*avgPriceMB*(1-demandMB) #4.072252

#Cross-elasticities
crossElasKR = -(beta1)*avgPriceKR*demandKR #1.012787 -> canibalization
crossElasMB = -(beta1)*avgPriceMB*demandMB #0.9574505
crossElasKB = -(beta1)*avgPriceKB*demandKB #0.9188917

#create a data frame that includes all the elasticities
elas_KB_KR_MB <- data.frame("Product"=c("KB","KR","MB"), "Own Elasticity"=c(ownElasKB,ownElasKR,ownElasMB), 
                            "Cross Price Elasticity"=c(crossElasKB,crossElasKR,crossElasMB))
elas_KB_KR_MB

##   Product Own.Elasticity Cross.Price.Elasticity
## 1      KB       4.257845              0.9054761
## 2      KR       4.131272              1.0199190
## 3      MB       4.069542              0.9601601

Part 1.3 - Profit Maximization and Optimal Price

Calculate optimal prices for KB and KR to maximize the profit when Mango price is priceMB = 1.43.

#Unit cost
uc=0.5
priceMB=1.43

demand_KB_KR=function(priceKB,priceKR,priceMB,para){
  probKB=exp(para[1]+para[4]*priceKB)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  probKR=exp(para[2]+para[4]*priceKR)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  return(cbind(probKB,probKR))
}

#define a profit function 
profit_KB_KR=function(priceKB,priceKR, priceMB,para){
  profitKB=1000*(demand_KB_KR(priceKB,priceKR, priceMB,para)[,1]*(priceKB-0.5))
  profitKR=1000*(demand_KB_KR(priceKB,priceKR, priceMB,para)[,2]*(priceKR-0.5))
  return(cbind(profitKB,profitKR))
}

#Choose space of prices to search for the optimal price over
#Because we search over two dimensions, create complete combination of the two prices
#Choose space of prices to search for the optimal price over
aux=seq(1,3,0.01)
#Because we search over two dimensions, create complete combination 
#of the two prices
pricespace=expand.grid(aux,aux)
#Compute profit at each realization of this price space.
#I write for-loop. While there are ways to get this done in a vectorized way,
#this for-loop code probably helps some in project 2.
#Calculate profit

#At each iteration of the loop, I take one realization of [P^KB,P^KR] pair and evaluate
#profit at that realization.
profitmat=matrix(0L,nrow(pricespace),1)

for (i in 1:nrow(pricespace)){
  profitmat[i]=sum(profit_KB_KR(pricespace[i,1],pricespace[i,2],1.43,para))  
}

#Visualize
xaxis=list(title="P^{KB}")
yaxis=list(autorange = "reversed",title="P^{KR}")
zaxis=list(title="Profit")
p=plot_ly(x=pricespace[,1],y=pricespace[,2],z=as.numeric(profitmat),
          type="scatter3d",mode="markers",
          marker = list(color = as.numeric(profitmat), colorscale = c('#FFE1A1', '#683531'), showscale = TRUE))%>%
  layout(scene=list(xaxis=xaxis,yaxis=yaxis,zaxis=zaxis))%>%
  config(mathjax = 'cdn')
p

#calculate max profit and optimal prices
max(profitmat) #393.408; optimal price for KB is 1.16, optimal price for KR is 1.16

## [1] 393.408

pricespace[profitmat==max(profitmat)]

## [1] 1.16 1.16

#create a data frame that contains optimal price and max profit information
infoTable = data.frame(maxProfit = max(profitmat), 
                       priceKR = pricespace[profitmat==max(profitmat)][1],
                       priceKB = pricespace[profitmat==max(profitmat)][2])
infoTable

##   maxProfit priceKR priceKB
## 1   393.408    1.16    1.16

Part 2 - Logit model with segmentation

Part 2.1 - Choose the number of segments

In this part, we are trying to study segmentation and understand how to position based on segmentation results. We use K-means to cluster customers and then decide a proper number for segments based on cluster analysis.

Insights: We decide to use 9 segments (including an additional segment for those who don’t fit in any cluster). As we can see from the segList shown above, when there are 10 segments, one of the segments contains less than 4 people (283 *0.014 = 3.962), which makes our segmentation less reliable. Therefore, we took a step back, and chose 9 as the number of our segments

#Clustering
#number of individuals in demo data
N=283
set.seed(0)
#create a copy of original data
data2 = data
#write a for-loop to test what is the reasonable number of segments
segList = list()

for(i in 1:18){
  demo_cluster = kmeans(x=demo[, 2:18], centers = i, nstart = 1000)
  data2=data
  
  # now combine cluster identity into the raw data
  cluster_id = data.frame(id = demo$id)
  cluster_id$cluster = demo_cluster$cluster
  data2 = merge(data2, cluster_id, by = "id", all.x = T)
  
  # for those who don't fit in any cluster, group them into one additional cluster
  data2$cluster[is.na(data2$cluster)] = i+1
  
  #segment share
  seg.share = c( table(demo_cluster$cluster),N - sum(table(demo_cluster$cluster))) / N
  
  #input seg.share into segTable
  segList[[i]] = seg.share
  
  #recover original data
  data2 = data2[,1:8]
}
segList

## [[1]]
## 1   
## 1 0 
## 
## [[2]]
##         1         2           
## 0.5795053 0.4204947 0.0000000 
## 
## [[3]]
##         1         2         3           
## 0.4734982 0.1307420 0.3957597 0.0000000 
## 
## [[4]]
##         1         2         3         4           
## 0.1837456 0.2473498 0.1236749 0.4452297 0.0000000 
## 
## [[5]]
##         1         2         3         4         5           
## 0.4240283 0.1236749 0.1448763 0.1201413 0.1872792 0.0000000 
## 
## [[6]]
##         1         2         3         4         5         6           
## 0.1978799 0.1060071 0.1872792 0.1448763 0.2438163 0.1201413 0.0000000 
## 
## [[7]]
##         1         2         3         4         5         6         7 
## 0.1342756 0.1872792 0.1201413 0.1378092 0.1095406 0.1660777 0.1448763 
##           
## 0.0000000 
## 
## [[8]]
##          1          2          3          4          5          6 
## 0.16607774 0.10954064 0.13780919 0.13427562 0.08127208 0.10954064 
##          7          8            
## 0.14134276 0.12014134 0.00000000 
## 
## [[9]]
##          1          2          3          4          5          6 
## 0.01413428 0.14840989 0.10247350 0.11307420 0.16961131 0.10954064 
##          7          8          9            
## 0.13427562 0.13780919 0.07067138 0.00000000 
## 
## [[10]]
##          1          2          3          4          5          6 
## 0.01413428 0.16961131 0.13427562 0.10954064 0.10247350 0.10247350 
##          7          8          9         10            
## 0.07773852 0.13780919 0.06713781 0.08480565 0.00000000 
## 
## [[11]]
##          1          2          3          4          5          6 
## 0.07773852 0.10600707 0.08833922 0.10247350 0.08480565 0.10954064 
##          7          8          9         10         11            
## 0.09893993 0.01413428 0.14840989 0.06713781 0.10247350 0.00000000 
## 
## [[12]]
##          1          2          3          4          5          6 
## 0.10600707 0.10954064 0.08480565 0.07420495 0.09893993 0.01413428 
##          7          8          9         10         11         12 
## 0.07773852 0.14840989 0.06713781 0.03533569 0.09540636 0.08833922 
##            
## 0.00000000 
## 
## [[13]]
##          1          2          3          4          5          6 
## 0.14840989 0.06360424 0.07067138 0.10954064 0.01413428 0.10247350 
##          7          8          9         10         11         12 
## 0.03533569 0.06713781 0.07773852 0.09540636 0.02473498 0.08833922 
##         13            
## 0.10247350 0.00000000 
## 
## [[14]]
##          1          2          3          4          5          6 
## 0.06360424 0.14840989 0.07067138 0.09893993 0.08127208 0.09540636 
##          7          8          9         10         11         12 
## 0.04240283 0.07773852 0.07773852 0.04946996 0.02120141 0.05300353 
##         13         14            
## 0.01413428 0.10600707 0.00000000 
## 
## [[15]]
##          1          2          3          4          5          6 
## 0.02120141 0.09540636 0.08833922 0.02473498 0.08833922 0.04240283 
##          7          8          9         10         11         12 
## 0.10247350 0.07773852 0.06713781 0.06360424 0.10954064 0.07773852 
##         13         14         15            
## 0.04593640 0.01413428 0.08127208 0.00000000 
## 
## [[16]]
##          1          2          3          4          5          6 
## 0.02473498 0.12014134 0.04946996 0.05653710 0.10247350 0.04240283 
##          7          8          9         10         11         12 
## 0.06713781 0.01413428 0.09540636 0.07420495 0.04593640 0.08833922 
##         13         14         15         16            
## 0.06007067 0.10600707 0.03180212 0.02120141 0.00000000 
## 
## [[17]]
##          1          2          3          4          5          6 
## 0.09540636 0.01413428 0.09893993 0.07420495 0.01766784 0.04240283 
##          7          8          9         10         11         12 
## 0.03180212 0.08127208 0.04593640 0.02120141 0.03180212 0.09893993 
##         13         14         15         16         17            
## 0.04946996 0.06713781 0.09540636 0.08480565 0.04946996 0.00000000 
## 
## [[18]]
##          1          2          3          4          5          6 
## 0.05653710 0.06007067 0.02120141 0.01413428 0.09187279 0.04946996 
##          7          8          9         10         11         12 
## 0.05653710 0.06713781 0.04946996 0.07773852 0.06713781 0.06360424 
##         13         14         15         16         17         18 
## 0.06007067 0.02120141 0.04240283 0.06360424 0.04240283 0.09540636 
##            
## 0.00000000

Now, we estimate separate demand models for each of the 9 segments.

demo_cluster = kmeans(x=demo[, 2:18], centers = 8, nstart = 1000)
summary(demo_cluster)

##              Length Class  Mode   
## cluster      283    -none- numeric
## centers      136    -none- numeric
## totss          1    -none- numeric
## withinss       8    -none- numeric
## tot.withinss   1    -none- numeric
## betweenss      1    -none- numeric
## size           8    -none- numeric
## iter           1    -none- numeric
## ifault         1    -none- numeric

# now combine cluster identity into the raw data
cluster_id = data.frame(id = demo$id)
cluster_id$cluster = demo_cluster$cluster
data = merge(data, cluster_id, by = "id", all.x = T)

# for those who don't fit in any cluster, group them into one additional cluster
data$cluster[is.na(data$cluster)] = 9

#segment share
seg.share = c( table(demo_cluster$cluster),N - sum(table(demo_cluster$cluster))) / N

#Estimate multinomial logit model separately for each segment
#store the coefficients
coef.est = data.frame(segment = 1:9, intercept.KB = NA, intercept.KR = NA, 
                      intercept.MB = NA, price.coef = NA) 

for (seg in 1:9) {
  # During each loop, pick subset of data of consumers from each segment.
  data.sub = subset(data, cluster == seg)
  
  #Using that data, the rest remains the same.
  mlogitdata = mlogit.data(data.sub,id="id",varying=4:7,choice="choice",shape="wide")
  
  #Run MLE.
  mle = gmnl(choice ~  price, data = mlogitdata)
  mle
  #Store the outcome in the coef.est matrix.
  coef.est[seg, 2:5] = mle$coefficients
}
coef.est

##   segment intercept.KB intercept.KR intercept.MB price.coef
## 1       1    3.9972969     3.958938    3.8837551  -3.715366
## 2       2    7.6063946     6.661934    7.4775392  -5.897474
## 3       3    4.8086045     4.606528    5.6294806  -4.517302
## 4       4    2.3336806     3.112574    2.9252012  -2.896447
## 5       5    7.3034174     7.138563    7.1181389  -5.793619
## 6       6    2.9694016     3.867674    2.7276100  -2.909001
## 7       7    3.8689983     4.354056    4.0523865  -3.502896
## 8       8    0.9169255     1.673183    0.4573439  -1.251711
## 9       9    5.1174301     4.509340    4.5449628  -4.062526

Part 2.2 - Compare Elasticities between Seg and Non-seg

Calculate own- and cross-elasticities with segmentation.

demand1=function(priceKB,priceKR,priceMB,para){
  probKB=exp(para[1]+para[4]*priceKB)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  probKR=exp(para[2]+para[4]*priceKR)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  probMB=exp(para[3]+para[4]*priceKR)/(1+exp(para[1]+para[4]*priceKB)+exp(para[2]+para[4]*priceKR)+exp(para[3]+para[4]*priceMB))
  return(cbind(probKB,probKR,probMB))
}
agg_choice=function(priceKB,priceKR,priceMB) {  #3 products segment
  
  agg_choice=seg.share[1]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[1,2:5]))+
    seg.share[2]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[2,2:5]))+
    seg.share[3]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[3,2:5]))+
    seg.share[4]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[4,2:5]))+
    seg.share[5]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[5,2:5]))+ 
    seg.share[6]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[6,2:5]))+
    seg.share[7]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[7,2:5]))+
    seg.share[8]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[8,2:5]))+
    seg.share[9]*demand1(priceKB,priceKR,priceMB,as.numeric(coef.est[9,2:5]))
  return(agg_choice)
}

avgPriceVec=list()
for (i in 1:3){
  avgPriceKB = mean(data$price.KB[data$cluster==i]) 
  avgPriceKR = mean(data$price.KR[data$cluster==i]) 
  avgPriceMB = mean(data$price.MB[data$cluster==i]) 
  avgPriceVec[[i]]=c(avgPriceKB,avgPriceKR,avgPriceMB)
}
###aggregate_choice and demand#########

para2=c(beta0KR,beta0MB,beta1)
demand2=function(price1,price2,para2){  #2 products, para contains 3 #s, func SAME NAME
  prob1=exp(para2[1]+para2[3]*price1)/(1+exp(para2[1]+para2[3]*price1)+exp(para2[2]+para2[3]*price2))
  prob2=exp(para2[2]+para2[3]*price1)/(1+exp(para2[1]+para2[3]*price1)+exp(para2[2]+para2[3]*price2))
  return(cbind(prob1,prob2))
}

agg_choice2=function(priceKB,priceKR, own, rival) {  #2 products segment
  m=c(own+1,rival+1,5)
  agg_choice=seg.share[1]*demand2(priceKB,priceKR,as.numeric(coef.est[1,m]))+
    seg.share[2]*demand2(priceKB,priceKR,as.numeric(coef.est[2,m]))+
    seg.share[3]*demand2(priceKB,priceKR,as.numeric(coef.est[3,m]))+
    seg.share[4]*demand2(priceKB,priceKR,as.numeric(coef.est[4,m]))+
    seg.share[5]*demand2(priceKB,priceKR,as.numeric(coef.est[5,m]))+ 
    seg.share[6]*demand2(priceKB,priceKR,as.numeric(coef.est[6,m]))+
    seg.share[7]*demand2(priceKB,priceKR,as.numeric(coef.est[7,m]))+
    seg.share[8]*demand2(priceKB,priceKR,as.numeric(coef.est[8,m]))+
    seg.share[9]*demand2(priceKB,priceKR,as.numeric(coef.est[9,m]))
  return(agg_choice)
}

#own elasticity 
elasticity_own=function(avgPriceAll,own){
  coef2=c(1,2,3)[c(1,2,3)!=own][1]
  coef3=c(1,2,3)[c(1,2,3)!=own][2]
  agg_choice=agg_choice(avgPriceAll[own],avgPriceAll[coef2],avgPriceAll[coef3])
  agg_choicenew=agg_choice(avgPriceAll[own]*1.01,avgPriceAll[coef2],avgPriceAll[coef3])
  (agg_choicenew-agg_choice)/agg_choice
  
} 
KB4=(elasticity_own(avgPriceAll,1)[1])*-100 #KB  -0.04392695 
KR4=(elasticity_own(avgPriceAll,2)[1])*-100 #KR   -0.04356302 
MB4=(elasticity_own(avgPriceAll,3)[1])*-100 #MB  -0.04112575

Own_elas_KB_KR_MB <- data.frame("Product"=c("KB","KR","MB"), "Own Elasticity"=c(KB4,KR4,MB4))
Own_elas_KB_KR_MB 

##   Product Own.Elasticity
## 1      KB       4.393557
## 2      KR       4.375514
## 3      MB       4.186238

Insights: With segmentation, the cross elasticities that KR and KB have on each other increased, meaning that KR and MB have become each other’s closer substitute compared to the non-segmentation case. In addition, MB has become a close substitute of KB (compared with the non-segmentation case that KR is the close substitute of KB), which implies that segmentation has helped to mitigate the risk of cannibalization between KB and KR.

#cross elasticity 
elasticity_cross=function(avgPriceAll,own){
  e=vector()
  coef2=c(1,2,3)[c(1,2,3)!=own][1]
  coef3=c(1,2,3)[c(1,2,3)!=own][2]
  agg_choice=agg_choice(avgPriceAll[own],avgPriceAll[coef2],avgPriceAll[coef3])[1]
  agg_choicenew1=agg_choice(avgPriceAll[own],avgPriceAll[coef2]*1.01,avgPriceAll[coef3])[1]
  agg_choicenew2=agg_choice(avgPriceAll[own],avgPriceAll[coef2],avgPriceAll[coef3]*1.01)[1]
  
  e=append(e, (agg_choicenew1-agg_choice)/agg_choice)
  e=append(e, (agg_choicenew2-agg_choice)/agg_choice)
  e
  
} 
Cros_KB_KR_MB=(elasticity_cross(avgPriceAll, 1))*100  
Cros_KB_KR_MB

## [1] 0.9459524 1.1224485

Cros_KR_KB_MB=(elasticity_cross(avgPriceAll, 2))*100 
Cros_KR_KB_MB

## [1] 0.9363006 1.1235926

Cros_MB_KB_KR=(elasticity_cross(avgPriceAll, 3))*100  
Cros_MB_KB_KR

## [1] 0.9351753 1.0089358

Part 2.3 - Segments preference & KB positioning

Insights: As we took the average of the difference between intercept.KB and intercept.KR and between intercept.KB and intercept.MB across segments, we can see that the absolute value of difference between MB and KB is smaller than that between KR and KB (0.0117 < 0.1067). This result explains the insight we get from the substitution pattern that MB is a closer substitute of KB compared to KR.

For segment 4, 6, 7, 8, the intercept.KR is bigger than intercept.KB, meaning that for these 4 groups, KR is more appealing to them than KB. Therefore, the firm should sell KR to segment 4, 6, 7, 8. For the rest of the segments (1, 2, 3, 5, 9), the intercept.KB is bigger than intercept.KR, meaning that KB is more appealing to these 5 segments than KR. Therefore, the firm should sell KB to segment 1, 2, 3, 5, 9.

popularityCompare = data.frame(segments = 1:9, 
                               differKB_KR = coef.est$intercept.KB-coef.est$intercept.KR, 
                               differKB_MB = coef.est$intercept.KB-coef.est$intercept.MB,
                               differMB_KR = coef.est$intercept.MB-coef.est$intercept.KR)
popularityCompare

##   segments differKB_KR differKB_MB differMB_KR
## 1        1  0.03835883   0.1135418 -0.07518293
## 2        2  0.94446055   0.1288554  0.81560515
## 3        3  0.20207671  -0.8208761  1.02295281
## 4        4 -0.77889343  -0.5915207 -0.18737278
## 5        5  0.16485417   0.1852785 -0.02042434
## 6        6 -0.89827253   0.2417916 -1.14006408
## 7        7 -0.48505759  -0.1833882 -0.30166938
## 8        8 -0.75625710   0.4595815 -1.21583864
## 9        9  0.60808976   0.5724673  0.03562241

differCompare = data.frame(avg_differKB_KR=mean(popularityCompare$differKB_KR),
                           avg_differKB_MB=mean(popularityCompare$differKB_MB))
differCompare

##   avg_differKB_KR avg_differKB_MB
## 1      -0.1067378       0.0117479

The blue points represent the segments that prefer KB more than KR, and they are less price sensitive. The red points represent the segments that prefer KR more than KB, and they are more highly price sensitive. The number of segments that prefer KB are the same with that prefer KR.

#Scatterplot of parameters - beta_0^{KB}-beta_0^{KR} against beta_1
plot(popularityCompare$differKB_KR, coef.est[,5], 
     main = ("Preference Comparison between KB and KR"),
     xlab="Preference of KB - Preference of KR",ylab=("Price Sensitivity"),
     xlim=c(-1.5,2),ylim=c(-6,-1),
     type='n')
points(popularityCompare$differKB_KR[4],coef.est$price.coef[4],cex=20*seg.share[4],col='red',pch=16)
points(popularityCompare$differKB_KR[6],coef.est$price.coef[6],cex=20*seg.share[6],col='red',pch=16)
points(popularityCompare$differKB_KR[7],coef.est$price.coef[7],cex=20*seg.share[7],col='red',pch=16)
points(popularityCompare$differKB_KR[8],coef.est$price.coef[8],cex=20*seg.share[8],col='red',pch=16)
points(popularityCompare$differKB_KR[1],coef.est$price.coef[1],cex=20*seg.share[1],col='blue',pch=16)
points(popularityCompare$differKB_KR[3],coef.est$price.coef[3],cex=20*seg.share[3],col='blue',pch=16)
points(popularityCompare$differKB_KR[5],coef.est$price.coef[5],cex=20*seg.share[5],col='blue',pch=16)
points(popularityCompare$differKB_KR[2],coef.est$price.coef[2],cex=20*seg.share[2],col='blue',pch=16)
points(popularityCompare$differKB_KR[9],coef.est$price.coef[9],cex=20*seg.share[9],col='blue',pch=16)

If we look at the two plots together, we can know that those who prefer KB between KB and KR are also the same people who prefer MB between KR and MB. We call these people the “bubble people”. It means that these segments do not necessarily prefer one brand over another. They simply prefer beverages with bubbles.

#Scatterplot of parameters - beta_0^{KR}-beta_0^{MB} against beta_1
plot(popularityCompare$differMB_KR, coef.est[,5],
     main = ("Preference Comparison between MB and KR"),
     xlab="Preference of MB - Preference of KR",ylab=("Price Sensitivity"),
     xlim=c(-1.5,2),ylim=c(-6,-1), 
     type='n')
points(popularityCompare$differMB_KR[4],coef.est$price.coef[4],cex=20*seg.share[4],col='red',pch=16)
points(popularityCompare$differMB_KR[6],coef.est$price.coef[6],cex=20*seg.share[6],col='red',pch=16)
points(popularityCompare$differMB_KR[7],coef.est$price.coef[7],cex=20*seg.share[7],col='red',pch=16)
points(popularityCompare$differMB_KR[8],coef.est$price.coef[8],cex=20*seg.share[8],col='red',pch=16)
points(popularityCompare$differMB_KR[2],coef.est$price.coef[2],cex=20*seg.share[2],col='blue',pch=16)
points(popularityCompare$differMB_KR[1],coef.est$price.coef[1],cex=20*seg.share[1],col='blue',pch=16)
points(popularityCompare$differMB_KR[3],coef.est$price.coef[3],cex=20*seg.share[3],col='blue',pch=16)
points(popularityCompare$differMB_KR[5],coef.est$price.coef[5],cex=20*seg.share[5],col='blue',pch=16)
points(popularityCompare$differMB_KR[9],coef.est$price.coef[9],cex=20*seg.share[9],col='blue',pch=16)

Part 2.4 - Optimal price and profit maximization (with and without KB)

Given that MB is priced at 1.43, suppose we only launch KR first. The optimal price of KR=1.07 and the maximal profit of KR=294.4419. In this case, profitMB = 109.1702

pricespace=seq(1,3,0.01)
uc=0.5
profit=1000*agg_choice2(pricespace,1.43,2,3)[,1]*(pricespace-uc)
priceKRBest=pricespace[profit==max(profit)] 
priceKRBest#best price for KR=1.07

## [1] 1.07

max(profit)  #max profit for KR 294.4419

## [1] 294.4419

Suppose next we do launch KB and now there are 3 products in the market. The optimal price of KB is 1.15 and the optimal price of KR is 1.19, with the max profit of Kiwi is 387.6119. In this case, profit of MB is 89.63174.

We can see that as Kiwi launches KB, profit of Kiwi increased to $93 and the profit of Mango decreased to $19.5.

The model justifies the launch of KB. From the perspective of consumer segmentation and product positioning, since beta0 represents the underlying popularity of a specific product, we can tell that the underlying popularity of KB is the highest among all of the 3 products in segment 1, 2, 5, 9, KR has the highest popularity in segment 4, 6, 7, 8 while MB has the highest popularity in segment 3 only(in which KB has the second highest popularity). Therefore, by selling KB to segment 1, 2, 5, 9, and KR to segment 4, 6, 7, 8, we can steal some market share from MB and also set a higher optimal price for KR (1.19) in this case. Now the inclusion of both KB and KR returns a higher profit of 387.61 than 294.44.

#profit for MB
profit_MB=1000*agg_choice2(1.43,priceKRBest,3,2)[,1]*(1.43-uc)  #109.1702  
profit_MB

##    prob1 
## 109.1702

#GIVEN MB, look at KR, KB
para=c(beta0KB,beta0KR,beta0MB,beta1)
profit_KB_KR_seg=function(priceKB,priceKR, priceMB,para){
  profitKB=1000*(agg_choice(priceKB,priceKR, priceMB)[,1]*(priceKB-0.5))
  profitKR=1000*(agg_choice(priceKB,priceKR, priceMB)[,2]*(priceKR-0.5))
  return(cbind(profitKB,profitKR))
}

aux=seq(1,3,0.01)
#Because we search over two dimensions, create complete combination 
#of the two prices
pricespace=expand.grid(aux,aux)

profitmat=matrix(0L,nrow(pricespace),1)
for (i in 1:nrow(pricespace)){
  profitmat[i]=sum(profit_KB_KR_seg(pricespace[i,1],pricespace[i,2],1.43,para))  
}

max(profitmat) #387.6119; 

## [1] 387.6119

priceKB_KRBest=pricespace[profitmat==max(profitmat)] #optimal price for KB is 1.15, optimal price for KR is 1.19
priceKB_KRBest

## [1] 1.15 1.19

#MB new profit

profit_MB2=1000*agg_choice(1.43,priceKB_KRBest,priceKB_KRBest)[2]*(1.43-uc)  #244.3255 new MB profit
profit_MB2 #89.63174

## [1] 89.63174

Part 3 - Understanding strategic responses when rival lowers its price

In reality, Mango might be as smart as you are in that they might also realize 1.43 might not be optimal price. This is to say, in the previous question, we assumed that Mango maintains its price at 1.43 regardless of what we do. Suppose now instead that they react and set an optimal price against our new prices, and we need to set an optimal prices against their new price, and so on.

Let’s see where this “pricing war” will be converged at and understand what is the strategic advantage of launching Kiwi Bubble now.

#round 1
#MANGO:
pricespace=seq(0,2,0.01)
uc=0.5
profit=1000*agg_choice(pricespace,1.15,1.19)[,1]*(pricespace-uc)

max(profit)  #170.9814

## [1] 170.9814

pricespace[profit==max(profit)] #0.96

## [1] 0.96

#KB KR
#Choose space of prices to search for the optimal price over
aux=seq(1,3,0.01)
#Because we search over two dimensions, create complete combination 
#of the two prices
pricespace=expand.grid(aux,aux)
profitmat=matrix(0L,nrow(pricespace),1)
for (i in 1:nrow(pricespace)){
  profitmat[i]=sum(profit_KB_KR_seg(pricespace[i,1],pricespace[i,2],0.96,para))  
}

max(profitmat) #269.6908

## [1] 269.6908

pricespace[profitmat==max(profitmat)] #1.01 1.09

## [1] 1.01 1.09

#round 2
#MANGO:
pricespace=seq(0,2,0.01)
uc=0.5
profit=1000*agg_choice(pricespace,1.01,1.09)[,1]*(pricespace-uc)

max(profit)  #140.0129

## [1] 140.0129

pricespace[profit==max(profit)] #0.92

## [1] 0.92

#KB KR
#Choose space of prices to search for the optimal price over
aux=seq(1,3,0.01)
#Because we search over two dimensions, create complete combination 
#of the two prices
pricespace=expand.grid(aux,aux)
profitmat=matrix(0L,nrow(pricespace),1)
for (i in 1:nrow(pricespace)){
  profitmat[i]=sum(profit_KB_KR_seg(pricespace[i,1],pricespace[i,2],0.92,para))  
}

max(profitmat) #256.2545

## [1] 256.2545

pricespace[profitmat==max(profitmat)] #1.00 1.08

## [1] 1.00 1.08

The “price war” ends at round 3 as the prices of both companies stay stable.

#round 3
#MANGO:
pricespace=seq(0,2,0.01)
uc=0.5
profit=1000*agg_choice(pricespace,1.00,1.08)[,1]*(pricespace-uc)

max(profit)  #137.4193

## [1] 137.4193

pricespace[profit==max(profit)] #0.92

## [1] 0.92

#KB KR
#Choose space of prices to search for the optimal price over
aux=seq(1,3,0.01)
#Because we search over two dimensions, create complete combination 
#of the two prices
pricespace=expand.grid(aux,aux)
profitmat=matrix(0L,nrow(pricespace),1)
for (i in 1:nrow(pricespace)){
  profitmat[i]=sum(profit_KB_KR_seg(pricespace[i,1],pricespace[i,2],0.92,para))  
}

max(profitmat) #256.2545

## [1] 256.2545

pricespace[profitmat==max(profitmat)] #1.00 1.08

## [1] 1.00 1.08

Conclusion:

After three rounds of “pricing war”, the market becomes stable and the prices for MB, KB and KR are 0.92, 1, and 1.08 respectively. By introducing KB, the amount of profit of KR shift to KB is the same as within-Kiwi substitution. KB steals MB’s market share by 21% and the profit increases by 40%.

In the case of the price war, revenue from KB is primarily coming from demand stealing from MB. This is in contrast to the no-price-war case, where stealing share accounts for less. Therefore, our original story that “Kiwi Bubbles should be positioned to steal demand from Mango with a low price” is reinforced in price-war environments.

Those who do not buy any in the “no response” scenario are buying MB in the “price war” scenario - hence KB launch induces more demand stealing. In both cases, Kiwi has to set a low price for KR if it does not launch KB - as KR is the only product, serving only the niche market is not optimal.

Limitations of the model: The formula for calculating cross elasticities suggests that a change in the price of a particular product has the same effect on the demand for any other products in the market. This is because the multinomial logit model is rather a simple model that does not allow so many complications. But in reality, a change in the price of a product should have a bigger effect on the demand of its close substitutes.